Determing size of aperture diameter on different focal lenghts lenses.

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I have always used the formula fl/f stop to determine the diameter of a lens opening. For example, a 55mm at 4.5 equals 12.2 mm diameter opening. A 75mm lens at f 6.15 also equals 12.2mm opeining. To anticipate the light on the gg for the 55mm lens, I took my 75mm at f6.1 to see the amount of light a 55mm would let in at f4.5. I was told by a reliable source this was not an accurate way to measure such. I am confused why it would not be. The only difference as I see it, is the angles at which the light strikes the gg. They will be slightly greater with the 55mm, but the center should still duplicate the same amount of light, right? Any input would be appreciated... Thank you...

-- Bill Glickman (bglick@pclv.com), February 24, 2000

Answers

Basically speaking, this is correct, although I understand that with some of the more complex optical formulas, this isnt always true. I will let others with a better understanding of optical design tell us why.

-- Ron Shaw (shaw9@llnl.gov), February 24, 2000.

I'm not sure I understand the question. The illumination on the ground glass at f/4.5 will be the same, whatever the focal length of the lens.

-- Alan Gibson (Alan@snibgo.com), February 25, 2000.

Alan, that is correct, I overlooked this simple fact. Thank you.

-- Bill Glickman (bglick@pclv.com), February 25, 2000.

One twist: point sources such as stars are imaged with a brighness that depends on the physical size of the aperture, not the f-number. Thus a 450 mm f9 will give a brighter image of Orion than a 150 mm f5.6, because the aperture diameter is 50 mm compared to 27 mm.

Knowing the definition of the f-number can also be useful when making Waterhouse stops for old lenses. The easy way to do this is to measure the brightness of an extended source (like a light box) on the ground glass and compare to a lens with an accurate iris. The 'proper' way is to measure the size of the hole as seen though the front element and divide by the focal length.

-- Struan Gray (struan.gray@sljus.lu.se), February 25, 2000.


The point-source thing is only true up to a point (pardon the pun). Diffraction comes into play, and the size of the Airey disc increases with numerical aperture. Your theoretically perfect f/5.6 lens will give a patch of light ~3.7 microns across, and the f/9 lens will spread its extra light energy over a circle of ~6 microns. So the difference in intensity is pretty negligible, (about 66/59 if my calculations are correct). So to get bright pictures of stars you need lenses that are both physically large, and have a low f number.

-- Pete Andrews (p.l.andrews@bham.ac.uk), February 28, 2000.


As a paid up member of pedants anonymous I feel obliged to point out that both Airey disks are effectively delta functions when convolved with the MTF of normal film. In laymans terms, most films cannot tell the difference between a 3.7 micron spot and a 6 micron spot, so the 450 f9 seems to give brighter stars even if it doesn't really.

I'll grudgingly admit that if you use a film like Tech Pan things start to look different, but there again, lenses that can cover LF and remain diffraction limited at f5.6 (or f9 for that matter) tend to hang out in spy planes, not camera stores.

-- Struan Gray (struan.gray@sljus.lu.se), February 29, 2000.


Well, you did shunt us off down this pedantic siding in the first place Struan. };^)

-- Pete Andrews (p.l.andrews@bham.ac.uk), February 29, 2000.

I'll take that as a compliment.

-- Struan Gray (struan.gray@sljus.lu.se), February 29, 2000.

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